1065. 字符串的索引对 🔒

题目描述

给出 字符串 text 和 字符串列表 words, 返回所有的索引对 [i, j] 使得在索引对范围内的子字符串 text[i]...text[j]（包括 i 和 j）属于字符串列表 words。

示例 1:

输入: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
输出: [[3,7],[9,13],[10,17]]

示例 2:

输入: text = "ababa", words = ["aba","ab"]
输出: [[0,1],[0,2],[2,3],[2,4]]
解释: 
注意，返回的配对可以有交叉，比如，"aba" 既在 [0,2] 中也在 [2,4] 中

提示:

所有字符串都只包含小写字母。
保证 words 中的字符串无重复。
1 <= text.length <= 100
1 <= words.length <= 20
1 <= words[i].length <= 50
按序返回索引对 [i,j]（即，按照索引对的第一个索引进行排序，当第一个索引对相同时按照第二个索引对排序）。

解法

方法一：暴力枚举

Python3JavaC++Go

class Solution:
    def indexPairs(self, text: str, words: List[str]) -> List[List[int]]:
        words = set(words)
        n = len(text)
        return [
            [i, j] for i in range(n) for j in range(i, n) if text[i : j + 1] in words
        ]

class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd = false;

    void insert(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }
}

class Solution {
    public int[][] indexPairs(String text, String[] words) {
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        int n = text.length();
        List<int[]> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            Trie node = trie;
            for (int j = i; j < n; ++j) {
                int idx = text.charAt(j) - 'a';
                if (node.children[idx] == null) {
                    break;
                }
                node = node.children[idx];
                if (node.isEnd) {
                    ans.add(new int[] {i, j});
                }
            }
        }
        return ans.toArray(new int[ans.size()][2]);
    }
}

class Trie {
public:
    vector<Trie*> children;
    bool isEnd = false;

    Trie() {
        children.resize(26);
    }

    void insert(string word) {
        Trie* node = this;
        for (char c : word) {
            c -= 'a';
            if (!node->children[c]) node->children[c] = new Trie();
            node = node->children[c];
        }
        node->isEnd = true;
    }
};

class Solution {
public:
    vector<vector<int>> indexPairs(string text, vector<string>& words) {
        Trie* trie = new Trie();
        for (auto w : words) trie->insert(w);
        int n = text.size();
        vector<vector<int>> ans;
        for (int i = 0; i < n; ++i) {
            Trie* node = trie;
            for (int j = i; j < n; ++j) {
                int idx = text[j] - 'a';
                if (!node->children[idx]) break;
                node = node->children[idx];
                if (node->isEnd) ans.push_back({i, j});
            }
        }
        return ans;
    }
};

type Trie struct {
    children [26]*Trie
    isEnd    bool
}

func newTrie() *Trie {
    return &Trie{}
}

func (this *Trie) insert(word string) {
    node := this
    for _, c := range word {
        idx := int(c - 'a')
        if node.children[idx] == nil {
            node.children[idx] = newTrie()
        }
        node = node.children[idx]
    }
    node.isEnd = true
}

func indexPairs(text string, words []string) [][]int {
    trie := newTrie()
    for _, w := range words {
        trie.insert(w)
    }
    n := len(text)
    var ans [][]int
    for i := range text {
        node := trie
        for j := i; j < n; j++ {
            idx := int(text[j] - 'a')
            if node.children[idx] == nil {
                break
            }
            node = node.children[idx]
            if node.isEnd {
                ans = append(ans, []int{i, j})
            }
        }
    }
    return ans
}

方法二：前缀树

相似题目：

Python3

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True


class Solution:
    def indexPairs(self, text: str, words: List[str]) -> List[List[int]]:
        trie = Trie()
        for w in words:
            trie.insert(w)
        n = len(text)
        ans = []
        for i in range(n):
            node = trie
            for j in range(i, n):
                idx = ord(text[j]) - ord('a')
                if node.children[idx] is None:
                    break
                node = node.children[idx]
                if node.is_end:
                    ans.append([i, j])
        return ans

1065. 字符串的索引对 🔒

题目描述

解法

方法一：暴力枚举

方法二：前缀树

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